Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2009 (C1.2, DMAC 2461, DMCO 2461, DMPC 2461)

This University of Oxford Part C paper combines Analytic Topology in section A and Axiomatic Set Theory in section B, cross-listed for Mathematics, Mathematics and Computer Science, and Mathematics and Philosophy. ^[ref] The archive retains the complete editable four-page transcription. Bounded official-source searches found no advertised answers for this exact paper. The solution to A1(iv) replaces a faulty printed radial hint with a proved rectangular construction.

Let $\mathcal{H}$ be the set of separately continuous real-valued functions on the plane. Take as a subbasis all sets $h^{-1}(U)$, where $h\in \mathcal{H}$ and $U$ is open in the usual real line. Unions of finite intersections of these sets form a topology, and every $h$ is continuous for it by construction. Conversely every topology making every $h$ continuous must contain the subbasis, hence this entire generated topology. Thus the required coarsest topology exists and is unique: it is the initial topology of the family $\mathcal{H}$.

The family is a set, being a subset of ${\mathbb{R}}^{{\mathbb{R}}^2}$, so this construction has no proper-class issue. The coordinate projections belong to $\mathcal{H}$; consequently this topology contains the ordinary Euclidean topology. This last observation will also make convergence in the new topology imply ordinary coordinatewise convergence.

Fix a horizontal line $L=\mathbb{R}\times \{c\}$. The restriction of every member of $\mathcal{H}$ to $L$ is continuous in its usual topology. Therefore every subbasic open set intersects $L$ in a usual open set, so the induced topology is no finer than the usual one. Conversely the first coordinate projection is continuous for the initial topology, and its inverse images induce every usual open set on $L$. The two topologies on $L$ are equal. The second projection gives the identical argument for a vertical line.

Write $F={\overline{Y}}^X$. For every nonzero rational $r$, the intersection of $Y$ with $\{r\}\times \mathbb{R}$ is dense in that vertical line, since the nonzero rational numbers are dense in the real line. Hence $\{r\}\times \mathbb{R}\subseteq F$. For each real $c$, it follows that $F\cap (\mathbb{R}\times \{c\})$ contains $(\mathbb{Q}\setminus \{0\})\times \{c\}$ and is closed in that line. This dense subset forces the whole horizontal line into $F$. Thus $F=X$: $Y$ is a countable dense subset. The two successive uses of line density are necessary; the initial topology is finer than the Euclidean topology, so Euclidean density alone would not suffice.

For a fixed nonzero second coordinate $c$, the rational expression defining $h(x,c)$ is continuous on the half-line where $xc>0$. On the opposite half-line it is constant. At $x=0$ the ratio tends to $1$, agreeing with the value on the other side. If $c=0$ the entire restriction is constant. By symmetry the restrictions with the first coordinate fixed are continuous too. Hence $h$ is separately continuous, and so is $h_t(x,y)=h(x-t,y-t)$ for every real $t$.

On the diagonal $D$, $h_t(t,t)=1$, while $h_t(s,s)=0$ for $s\ne t$. ThereforeEach diagonal singleton is open in the induced topology; arbitrary unions show that $D$ has the discrete topology.

If $X$ were metrizable, its countable dense subset from the preceding part would make it second countable. Indeed balls with rational radii about dense points contain a countable basis after the usual smaller-ball argument. Every subspace of a second-countable space is second countable. A discrete second-countable space is countable, since each singleton must itself occur as a basis member. But $D$ is uncountable. This contradiction proves $X$ is not metrizable.

Suppose for contradiction that $p_n=(u_n,v_n)\in Y$ converges to the origin in the initial topology. The coordinate projections then give $u_n\to 0$ and $v_n\to 0$ in the usual real line. Rectangular bumps detecting nonconvergence, using anisotropic supports, avoids any unsupported regularity assertion about a radial maximum. SetThe radii are positive since neither coordinate of $p_n$ vanishes. Each rectangular support misses both coordinate axes, so $B$ vanishes on those axes. For a fixed $y=c\ne 0$, all but finitely many vertical support intervals miss $c$, because their centres and radii tend to zero. Hence $B(x,c)$ is a finite maximum of continuous functions of $x$, and is continuous. At $c=0$ it is identically zero. The same argument with the coordinates exchanged proves separate continuity everywhere. Thus $B$ is continuous for the initial topology.

However $B(p_n)=1$ for every $n$, whereas $B(0,0)=0$. This contradicts preservation of convergent sequences by a continuous function. Therefore no sequence from $Y$ converges to the origin.

The radial hint in the scan is not separately continuous for every candidate sequence: for $p_n=(2^{-n},2^{-2n})$, its value at the origin is zero, while its value at $(2^{-n},0)$ is at least $1-2(2^{-2n})/\sqrt{2^{-2n}+2^{-4n}}$, which tends to one. The rectangular construction supplies the needed valid proof independently of that hint.

If the origin had a countable local basis $U_1,U_2,\dots$, replace it by the decreasing local basis $V_n={\bigcap }_{j\le n}{U}_j$. Each $V_n$ meets the dense set $Y$, so choose $p_n\in V_n\cap Y$. Given any neighbourhood $U$ of the origin, some $U_j$ is contained in $U$; for every $n\ge j$ we then have $p_n\in V_n\subseteq U_j\subseteq U$. Thus $p_n$ converges to the origin, contradicting the preceding part. Consequently first countability fails already at the origin, and $X$ is not first countable.

Urysohn's lemma states that if $X$ is normal and $A,B$ are disjoint closed subsets, there is a continuous function $u:X\to [0,1]$ with $u{|}_A=0$ and $u{|}_B=1$. The values on the rest of $X$ remain in that interval. Interchanging $A,B$ or composing with an affine function gives any two prescribed real endpoint values. This is functional separation of closed sets, stronger than merely choosing disjoint open neighbourhoods.

A C-embedded subset $Y\subseteq X$ is one when every continuous real-valued function on $Y$ extends continuously to $X$. It is a Cb-embedded subset when this is required for every bounded continuous real-valued function, with a bounded extension. These definitions concern the subspace topology on $Y$.

Here is the bounded Tietze extension construction from Urysohn's lemma. Let $g:Y\to \mathbb{R}$ be continuous with $|g|\le M$. The case $M=0$ has the zero extension. For $M>0$, the sets $A=\{g\le -M/3\}$ and $B=\{g\ge M/3\}$ are disjoint and closed in $X$, since $Y$ is closed. Urysohn's lemma gives a continuous $u_0:X\to [-M/3,M/3]$ taking the endpoint values on $A,B$. On $Y$,Indeed on $A$ or $B$ this follows from the prescribed endpoint; on their complement both $|g|$ and $|u_0|$ are at most $M/3$.

Repeat on the residual. With $M_n=M(2/3{)}^n$, choose continuous $u_n$ satisfying $\Vert u_n{\Vert }_{\infty }\le M_n/3$ and reducing the residual on $Y$ to at most $M_{n+1}$. The series $F={\sum }_{n\ge 0}{u}_n$ converges uniformly by the geometric majorant. Its sum is continuous, has $|F|\le \sum M_n/3=M$, and restricts to $g$, since the residual tends uniformly to zero. Thus every bounded continuous function extends, with its bound preserved, proving $C^b$-embeddedness without assuming the extension theorem being proved.

For a strict-range modification of a continuous extension, the closed sets $Y$ and $Z=\{x:|f(x)|\ge 1\}$ are disjoint. Choose by Urysohn's lemma a continuous $u:X\to [0,1]$ with $u{|}_Y=0$ and $u{|}_Z=1$, and defineThis function is continuous and agrees with $f$ on $Y$. On $Z$ it vanishes. Off $Z$ we have $|f|<1$ and $0\le 1-u\le 1$, so $|\tilde{f}|\le |f|<1$. Hence its range lies strictly inside $(-1,1)$ on all of $X$. The construction also covers empty $Y$ or empty $Z$ by a constant separating function where appropriate.

Let $h:Y\to \mathbb{R}$ be continuous, without a boundedness assumption. The homeomorphism $\theta (t)=t/(1+|t|)$ maps the real line onto $(-1,1)$, with inverse ${\theta }^{-1}(s)=s/(1-|s|)$. The bounded function $g=\theta \circ h$ extends by the preceding bounded construction to a continuous real-valued $G$ on $X$. Since $|G|<1$ on $Y$, the strict-range modification gives a continuous $\tilde{G}$ with range in $(-1,1)$ and the same restriction to $Y$. Thereforeis continuous everywhere: its denominator is pointwise positive. On $Y$ it equals ${\theta }^{-1}(\theta (h))=h$. Thus $Y$ is $C$-embedded. It is important to modify the extension before applying the inverse: an extension allowed to attain $1$ or $-1$ outside $Y$ would make this formula undefined there.

Take the natural numbers as the dense discrete subspace of their Stone-Čech compactification $\beta \mathbb{N}$. The ambient compact Hausdorff space is normal, and the Stone-Čech extension property makes this subspace $C^b$-embedded. It is not $C$-embedded: the continuous function $n\mapsto n$ is unbounded, whereas every continuous real-valued function on a compact space is bounded. The subspace is not closed, so this does not contradict the preceding closed-subspace theorem.

A filter base on a set $X$ is a nonempty family $\mathcal{F}$ of nonempty subsets such that for every $F,G\in \mathcal{F}$ there is $H\in \mathcal{F}$ with $H\subseteq F\cap G$. Repeated application gives a member contained in any specified finite intersection. In particular every finite intersection is nonempty, so the family has the finite intersection property. Its upward closure $\{A\subseteq X:\exists F\in \mathcal{F},F\subseteq A\}$ is the corresponding proper filter.

For a connected intersection of a closed filter base, the filter-base property implies the finite intersection property. Compactness, in its closed-set formulation, therefore gives $K=\bigcap \mathcal{F}\ne \varnothing$. It is closed in $X$, hence compact.

Suppose additionally that each base member is connected but $K$ is disconnected. Write $K=A\cup B$ with $A,B$ disjoint nonempty closed subsets of $K$. They are compact, hence closed in the Hausdorff space $X$. Normality of a compact Hausdorff space gives disjoint open sets $U,V$ containing $A,B$, respectively. The compact set $W=X\setminus (U\cup V)$ misses $K$. For every $w\in W$ some $F_w\in \mathcal{F}$ misses $w$, so the open sets $X\setminus F_w$ cover $W$. A finite subcover gives $F_1\cap \cdots \cap F_m\subseteq U\cup V$. Choose $F_0\in \mathcal{F}$ inside this finite intersection. If $W$ is empty, choose any base member instead.

Since $F_0$ contains $K$, it meets both $U$ and $V$; but it lies in their disjoint union. These two relatively open nonempty pieces disconnect $F_0$, a contradiction. Thus the intersection is nonempty and connected. Compactness is used to pass from an infinite exclusion condition to a single base member.

A Hausdorff compactification of $X$ is an embedding $h:X\to Y$ into a compact Hausdorff space with dense image. The complement $Y\setminus h[X]$ is the remainder; a one-point or two-point compactification has a remainder of that cardinality.

For one point, take the unit circle andThe angle runs over $(-\pi ,\pi )$, so this is a homeomorphism onto the circle with $(-1,0)$ removed. That image is dense, and the circle is compact Hausdorff. For two points, take $Y=[-\pi /2,\pi /2]$ and $h_2(x)=\arctan x$. This is a homeomorphism onto its dense open interval, leaving exactly the two endpoints. Therefore both required compactifications exist; in the first the positive and negative ends meet, whereas in the second they remain distinct.

Write $T_n=h(\{x\in \mathbb{R}:|x|>n\})$, making explicit that these are subsets of $Y$. If $p\notin h[\mathbb{R}]$ and $U$ is an open neighbourhood of $p$, then $h([-n,n])$ is compact and hence closed in $Y$. The open set $U\setminus h([-n,n])$ still contains $p$ and meets the dense image of $h$. Such an intersection point belongs to $T_n$, so $p\in \overline{T_n}$ for every $n$.

Conversely, if $p=h(x)$ choose an integer $n>|x|$. Since $h$ is an embedding, there is an open neighbourhood $U$ of $p$ in $Y$ whose intersection with $h[\mathbb{R}]$ is $h((-n,n))$. It misses $T_n$, so $p\notin \overline{T_n}$. HenceFor the two ends of a real-line compactification, let $A_n^{+}=\overline{h((n,\infty ))}$ and $A_n^{-}=\overline{h((-\infty ,-n))}$. Each is nonempty compact and connected, being the closure of a connected image. The two nested families therefore have nonempty connected intersections $C^{+}$ and $C^{-}$. Since closure commutes with finite unions, $\overline{T_n}=A_n^{+}\cup A_n^{-}$. MoreoverFor if a point fails both intersections, it fails one positive member and one negative member, hence both members at the larger index by nesting.

If the remainder were a finite three-point set, its Hausdorff subspace topology would be discrete. Each nonempty connected subset $C^{\pm }$ would then be a singleton, so their union would have at most two points. This contradiction proves there is no three-point Hausdorff compactification of the real line. The argument does not assert that every possible remainder is finite.

The cumulative hierarchy is defined by transfinite recursion on the ordinals:The recursion theorem of ZF supplies each set stage. The whole universe is $V={\bigcup }_{\alpha \in \mathrm{Ord}}{V}_{\alpha }$, a proper-class union. Equivalently, $x\in V_{\alpha }$ exactly when the rank of $x$ is less than $\alpha$, where $\mathrm{rank}(x)=\sup \{\mathrm{rank}(z)+1:z\in x\}$.

A set $T$ is transitive if every element of an element of $T$ is itself in $T$, equivalently each element of $T$ is a subset of $T$. Proceed by transfinite induction. The empty set is transitive. If $T$ is transitive and $u\in v\in \mathcal{P}(T)$, then $v\subseteq T$ gives $u\in T$, and transitivity of $T$ gives $u\subseteq T$. Thus $u\in \mathcal{P}(T)$, proving the successor stage transitive. A union of transitive sets is transitive: if $u\in v\in {\bigcup }_i{T}_i$, choose $i$ with $v\in T_i$ and infer $u\in T_i$. This proves the limit step and hence every $V_{\alpha }$ is transitive.

For $x\in V_{\alpha }$, transitivity of $V_{\alpha }$ gives $x\subseteq V_{\alpha }$. Therefore $x\in \mathcal{P}(V_{\alpha })=V_{\alpha +1}$, provingTogether with the union definition at limits, transfinite induction now gives $V_{\alpha }\subseteq V_{\beta }$ whenever $\alpha \le \beta$. No rank theorem is needed for this monotonicity argument.

Induct on $\alpha$. For every $\beta <\alpha$, the induction hypothesis says $\beta \in V_{\beta +1}$, and monotonicity gives $V_{\beta +1}\subseteq V_{\alpha }$. Thus every element of the ordinal $\alpha$ lies in $V_{\alpha }$, so $\alpha \subseteq V_{\alpha }$. Taking its power set givesFor $\alpha =0$ the inclusion is vacuous and $\varnothing \in \mathcal{P}(\varnothing )$ supplies the base case. This argument uses ordinal membership as strict order and the previously proved hierarchy monotonicity.

Take a nonzero limit ordinal $\lambda$ and $x\in V_{\lambda }$. Choose $\beta <\lambda$ with $x\in V_{\beta }$. Transitivity gives $x\subseteq V_{\beta }$. Hence every actual subset of $x$ belongs to $\mathcal{P}(V_{\beta })=V_{\beta +1}$, and the actual set $\mathcal{P}(x)$ belongs to $V_{\beta +2}$. Since $\lambda$ is a limit, $\beta +2<\lambda$, so $\mathcal{P}(x)\in V_{\lambda }$. Thus the internal power set contains every actual subset of $x$, and the power-set axiom holds in $V_{\lambda }$.

Here the absoluteness fact used is: every bounded formula is absolute between transitive sets or classes with the actual membership relation. In particular $z\subseteq x$, expressed by $\forall t\in z(t\in x)$, has the same meaning internally and externally. The argument supplies the actual power set as an internal witness. If zero is included in one's convention for limit ordinals, its empty stage has no nonempty first-order domain; the assertion is understood for nonzero limits.

If also $\lambda >\omega$, the preceding ordinal-membership result gives $\omega \in V_{\omega +1}\subseteq V_{\lambda }$. The actual set $\omega$ contains the empty set and is closed under $z\mapsto z\cup \{z\}$. These bounded membership and successor conditions are absolute in the transitive stage, so $\omega$ is an internal inductive set. Therefore Infinity holds as well. In contrast $V_{\omega }$ has only hereditarily finite sets and cannot use $\omega$ as an element.

To prove replacement failure at V omega plus omega, put $M=V_{\omega +\omega }$. The set $\omega$ belongs to $M$, and for every $n\in \omega$ the ordinal $\omega +n$ also belongs to $M$. The relation saying that $y$ is obtained by taking $n$ successive ordinal successors of $\omega$ is definable, with parameter $\omega$, and is single valued in $M$.

For clarity this does not assume Replacement inside $M$. For each finite $n$ its finite computation sequence exists in $M$: its entries have rank below $\omega +n+1$, and the finite coding overhead still has rank below $\omega +\omega$. The bounded formula verifying this finite successor chain is absolute between $M$ and the universe. The ordinary finite induction identifying its unique last entry uses the actual finite ordinals, which are the same in $M$. This justifies the internal definable function $n\mapsto \omega +n$ without invoking the axiom being tested.

If Replacement held there, its range $B=\{\omega +n:n\in \omega \}$ would be an element of $M$. But ordinal rank equals the ordinal, soEvery element of $V_{\omega +\omega }$ has strictly smaller rank. This contradiction proves Replacement fails in $V_{\omega +\omega }$. Each individual output is present; the missing object is the set collecting all outputs.

König's theorem for cardinal numbers states that, for a set-indexed family with ${\kappa }_i<{\lambda }_i$ at every index,Here the sum is the cardinality of a disjoint union and the product is the cardinality of a Cartesian product. Work in ZFC. Choose disjoint sets $A_i$ of sizes ${\kappa }_i$ and sets $B_i$ of sizes ${\lambda }_i$. For any function $f:{⨆}_i{A}_i\to {\prod }_i{B}_i$, its $i$th coordinate values on $A_i$ form a subset of $B_i$ of cardinality at most ${\kappa }_i$. Choose $b_i\in B_i$ outside that subset, using choice. Then $b=(b_i{)}_i$ cannot equal $f(a)$ for any $a$: if $a\in A_j$, its $j$th coordinate was specifically excluded. Thus there is no surjection from the disjoint union onto the product.

Cardinal comparability under choice now implies the displayed strict inequality. Empty index set gives $0<1$ directly; empty individual $A_i$ cause no difficulty, since the corresponding coordinate excludes an empty image. This is the cardinal diagonal theorem, rather than the finitely branching tree result also bearing König's name.

The generalized continuum hypothesis isEquivalently, for every ordinal $\alpha$, $2^{{\aleph }_{\alpha }}={\aleph }_{\alpha +1}$. Here $2^{\kappa }$ is the cardinality of the power set of a set of cardinality $\kappa$, and ${\kappa }^{+}$ is the least strictly larger cardinal. The hypothesis says that no intermediate cardinal lies between an infinite set and its power set.

For GCH arithmetic at aleph omega, let $\kappa ={\aleph }_{\omega }$. Use monotonicity of cardinal exponentiation, the law $(a^b{)}^c=a^{bc}$, and $\kappa \cdot \kappa =\kappa$ for infinite cardinals in ZFC. Since $2\le \kappa \le 2^{\kappa }$,Thus ${\kappa }^{\kappa }=2^{\kappa }$, and the generalized continuum hypothesis yieldsThe lower bound embeds binary-valued functions into $\kappa$-valued ones; the upper bound codes each value by a subset and combines the resulting pairs of indices. In particular the equality before applying the hypothesis is a ZFC equality.

Write again $\kappa ={\aleph }_{\omega }$. The countable cardinal sum ${\sum }_{n<\omega }{\aleph }_n$ equals $\kappa$: it is at least their supremum, and at most ${\aleph }_0\cdot \kappa =\kappa$. König's theorem givesThe latter product injects into $P={\prod }_{n<\omega }{\aleph }_n$ by fixing coordinate zero and putting its $n$th coordinate in coordinate $n+1$. Conversely $P$ injects into the set of all functions $\omega \to \kappa$. ThereforeNo cardinal lies strictly between $\kappa$ and its successor, soThe strict lower bound is essential: replacing the product by the supremum of its factors would incorrectly give ${\aleph }_{\omega }$.

The product from the preceding part injects into ${\kappa }^{{\aleph }_0}$, where $\kappa ={\aleph }_{\omega }$, because each coordinate value is an ordinal below $\kappa$. Combining its strict König lower bound with the exponentiation upper bound givesHenceOnly the instance $2^{{\aleph }_{\omega }}={\aleph }_{\omega +1}$ of the generalized continuum hypothesis was needed in these three computations, though the question assumes the full hypothesis. No assumption that countable exponentiation leaves this singular cardinal unchanged is valid: König's theorem rules that out.

Let $U_1\subseteq U_2$ be transitive classes, both interpreted with actual membership. A formula $\varphi (\bar{x})$ is upward absolute from $U_1$ to $U_2$ if for every tuple of parameters $\bar{a}$ from $U_1$,For proper classes this means the usual relativization of the fixed formula's quantifiers. Full absoluteness requires the implication in both directions. The qualification on parameters matters: the smaller class must contain every object named in the comparison.

A sufficient syntactic condition is that $\varphi$ be a sigma-one formula in set theory, written ${\Sigma }_1$: an existential block $\exists z_1\cdots \exists z_m\theta$ with $\theta$ bounded, meaning that each of its quantifiers is of the form $\forall t\in a$ or $\exists t\in a$. A bounded formula is also called ${\Delta }_0$ and is absolute in both directions between transitive classes.

Prove the latter fact by structural induction. Atomic membership and equality have their actual meanings in both classes, and Boolean connectives preserve equivalence of truth. For a bounded quantifier with bound $a\in U_1$, transitivity ensures that every element of $a$ belongs to $U_1$; thus its range is exactly the same in both classes, and induction applies to its matrix. This proves the bounded case. If the smaller class satisfies a ${\Sigma }_1$ formula, take its existential witnesses in $U_1$. They remain available in $U_2$, and bounded absoluteness preserves the matrix. Hence ${\Sigma }_1$ formulas are upward absolute. An arbitrary unbounded universal quantifier has no such witness-preservation argument.

Use the bounded formulaIt expresses $x\subseteq y$ with no unbounded quantifier. For parameters $x,y\in U_1$, transitivity puts every member of $x$ in $U_1$ already, and membership in $y$ has the same meaning in both classes. Thus the formula is ${\Delta }_0$ and is absolute in both directions, in particular upward absolute.

Equality with the union is expressed byThe first conjunct says that every member of $x$ lies in some member of $y$; the second gives the reverse inclusion. Every quantifier is bounded by a parameter or by an element of a previously bounded parameter. Therefore bounded absoluteness applies, proving full absoluteness and hence upward absoluteness of $x=\bigcup y$. This only compares the relation for existing parameters; it does not require either class to contain the union of every one of its sets.

A bounded expression for the successor relation isThe first two clauses put $y$ and all its elements into $x$, while the last forbids any additional member of $x$. Extensionality identifies this exactly with $x=y\cup \{y\}$. Since the formula is ${\Delta }_0$, it is absolute in both directions between the transitive classes and therefore upward absolute. This bounded description also justifies the successor computations used earlier in the rank-stage Replacement counterexample.

To prove that power-set equality is not upward absolute, use the standard constructible universe theorem: $L$ is a transitive inner model of ZF containing every ordinal, and set rank is absolute between $L$ and $V$. Since $V\ne L$, some rank stage contains a nonconstructible set. Let $\alpha$ be the least ordinal with $V_{\alpha }⊈L$. It is not zero; it cannot be a limit, since a union of earlier stages contained in $L$ is still contained in $L$. Thus $\alpha =\beta +1$ and $V_{\beta }\subseteq L$.

We need not merely this inclusion, but that $y=V_{\beta }$ itself belongs to $L$. Inside $L$, ZF constructs its own rank stage $(V_{\beta }{)}^L$, which is a set of $L$. By transitivity and rank absoluteness,Consequently $y\in L$. Let $x=(\mathcal{P}(y){)}^L$, the power set computed in $L$. Then $x\in L$ and $L\models x=\mathcal{P}(y)$.

By minimality there is $z\in V_{\beta +1}\setminus L$. Externally $z\subseteq V_{\beta }=y$, but $z\notin x$, because every member of the set $x\in L$ is in the transitive class $L$. Hence $V/\models x=\mathcal{P}(y)$. Taking $U_1=L$ and $U_2=V$ proves power-set equality is not upward absolute, even between transitive models of ZF.

There is no contradiction with absoluteness of $z\subseteq y$: that bounded relation agrees whenever both parameters are present in the smaller class. Power-set equality quantifies over all subsets, including a new $z$ absent from the smaller class. The step proving $V_{\beta }\in L$ is essential; inclusion in a class alone would not imply membership in it.